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tigry1 [53]
3 years ago
5

G use the margin of error, confidence level, and standard deviation σ to find the minimum sample size required to estimate an un

known population mean μ. margin of error: $121, confidence level: 95%, σ = $528
Mathematics
1 answer:
kodGreya [7K]3 years ago
4 0
The sample size should be 73.

We use the formula

n=(\frac{z*\sigma}{E})^2

To find the z-score:
Convert 95% to a decimal:  0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this value goes with a z-score of 1.96.  Using this, our value for σ and our margin of error, we have:
n=(\frac{1.96(528)}{121})^2=73.15\approx73
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Step-by-step explanation:

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3 years ago
Shureka Washburn has scores of 67​, 68​, 76​, and 63 on her algebra tests. a. Use an inequality to find the scores she must make
mash [69]
Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
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solving the above we get:
274+2x≥71×6
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b] The above answer is x≥76, the mean of this is that if Shureka is aiming at getting an average of 71 or above, then she should be able to get a minimum score of 76 or above. Anything less than 76 will drop her average lower than 71.


8 0
3 years ago
What is the correct answer can you guys help me I really need help on this?? Please
luda_lava [24]
The answer is 192 y^8.

(4y^2)^3 = 64y^6

64y^6 • 3y^2= 192y^8

When “multiplying” a number with an exponent, you add the 2 exponents.
3 0
3 years ago
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